Leetcode 119

I will explain the first solution.

If you want to understand the second solution click on the link above the code.

So the first solution only wants us to return the last row of the triangle.

Since we need two rows always for this problem the current row and the previous row. In this problem res is the current row and temp is the previous row.

Here is how the problem works

We have res[0] equal to 1 so all we have to do is keep on appending 1 to res every time we iterate.
Then we have to loop through res’s indexes starting at 1 and ending at len(res) - 1 because we don’t want to count the start and end 1’s.
Inside the nested for we can make res[j] = temp[j-1] + temp[j]. If you don’t understand why this works look at the image bellow:
After iterating through everything we can return res.

The Code: This is the one that is 100%, 100%

func getRow(rowIndex int) []int {
    res := []int{}

    for i := 0; i <= rowIndex; i++ {
        temp := make([]int, len(res))
        copy(temp, res)
        res = append(res, 1)
        for j := 1; j < len(res)-1; j++ {
            res[j] = temp[j-1] + temp[j]
        }
    }
    return res
}

We could do somthing similar to what we did in Leetcode 118 But it is a pretty brute force solution for this problem:

func getRow(rowIndex int) []int {
    res := [][]int{}

    for i := 0; i < rowIndex; i++ {
        temp := make([]int, i+1)
        temp[0], temp[len(temp)-1] = 1, 1
        for j := 1; j < len(temp)-1; j++ {
            temp[j] = res[i-1][j-1] + res[i-1][j]
        }
        res = append(res, temp)
    }

    return res[len(res) - 1]
}